솔루션피아

$$\begin{cases} y_0 &= 12\ut{m}\\ v_x &= 2.50\ut{m/s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1=0.900\ut{sec}\\ t_2 : \text{Finish}\\ \end{cases}$$ (a) $\Delta x_1=?$ $$\begin{aligned} \Delta x &= v_xt,\\ &=2.25\ut{m} \end{aligned}$$ (b) $y_1=?$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y &= v_{..

$$\begin{cases} y_0 &= 2.42\ut{m}\\ v_x &= 23.6\ut{m/s}\\ h_N &= 0.90\ut{m}\\ \Delta x &= 12\ut{m} \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1 : \text{On Net}\\ \end{cases}$$ (a),(b) $\text{horizon}$ $$\Delta x = v_x t,$$ $$\therefore t = \frac{\Delta x}{v_x}$$ $$\begin{aligned} S &= v_0t+\frac{1}{2}at^2,\\ \..

$$\begin{cases} \Delta x &= 45.7\ut{m}\\ v_0 &= 460\ut{m/s}\\ \Delta y &= 0 \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1 : \text{Finish}\\ \end{cases}$$ $$\begin{aligned} \Delta x &= v_x t,\\ &= v_0t\cos\theta\\ \end{aligned}$$ $$t=\frac{\Delta x}{v_0\cos\theta}$$ $$\begin{aligned} S &= v_0 t+ \frac{1}{2}at^2,..

$$\begin{cases} v_0 &= 30.0\ut{m/s}\\ \theta_0 &= 40.0\degree\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1 : \text{Vertical Max}\\ t_2 : \text{Vertical Max after half point}\\ \end{cases}$$ (a) $v_1=?$ $$\begin{aligned} v_1 &= v_x\\ &= v_0\cos\theta_0\\ &= 30\cos40\degree\\ &\approx 22.9813332936\ut{m/s}\\ &..

$$\begin{cases} A : \text{Plane}\\ B : \text{Object} \end{cases}$$ $$\begin{cases} \theta_0 &= 52.0\degree\\ v_A &= \text{Const.}\\ h_0 &= 720\ut{m}\\ \Delta t &= 6.00\ut{s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ (a) $v_A=?$ $$\Delta r = v_{0}t+\frac{1}{2}at^2, $$ $$\begin{aligned} \Delta y_B &= v_{By0}t+\frac{1}{2}a_{By}t^2\\ -720 &= v_{By0}..

(풀이자주 : 속력이 m단위라 임의로 m/s로 간주하여 풀었습니다.) $$\begin{cases} \theta_0 &= 30\degree\\ v_0 &= 25.0\ut{m/s}\\ d &= 20.0\ut{m}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \Delta t_{0\rarr1} &= \frac{\Delta x_{0\rarr1}}{v_x}\\ &=\frac{d}{v_0\cos\theta}\\ &=\frac{8}{5 \sqrt{3}} \end{aligned}$$ (a) $h_1=?$ $$\Delta r = v_{0}t+\frac{1}{2}at^2,..

$$\begin{cases} \vec r_Q &= (32,3.5)\ut{m}\\ \vec r_P &= (0,1.0)\ut{m}\\ \vec v_0 &= (18\ut{m/s},40\degree)\\ \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\title{Over Q Point Possible?}$$ $$v_0 = 18\cos40\degree\i+18\sin40\degree\j,$$ $$\begin{aligned} t &= \frac{\Delta x}{v_x}\\ &= \frac{32}{18\cos40\degree}\\ &= \frac{16}{9} \sec40\degree\\ \end{aligned}$$ $$\begin{aligned} \..

$$\begin{cases} \vec v_{A0} &= (23.1\ut{m/s},45\degree)\\ x_{B0} &= 55\ut{m}\\ \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\bar v_B=?$$ $$\begin{aligned} \vec v_{A0} &= (23.1\ut{m/s},45\degree)\\ &= \frac{231}{10 \sqrt{2}}\i+\frac{231}{10 \sqrt{2}}\j\\ \end{aligned}$$ $$\begin{aligned} \Delta y_A&=v_{Ay0}t+\frac{1}{2}a_yt^2,\\ 0 &= v_{Ay0}t+\frac{1}{2}(-g)t^2\\ &= v_{Ay0}-\fra..

$$\begin{cases} v_0 &= 6.00v_{H}\\ \vec a &= -g\j\\ \end{cases}$$ $$\theta_0=?$$ $$\begin{aligned} \vec v_H &= \vec v_x,\\ \end{aligned}$$ $$\begin{aligned} \theta_0&=\cos^{-1}\frac{v_x}{v_0}\\ &=\cos^{-1}\frac{v_H}{6v_H}\\ &=\cos^{-1}\frac{1}{6}\\ &\approx 1.40334824758\ut{rad}\\ &\approx 1.40\ut{rad}\\ \end{aligned}$$

$$\begin{cases} \vec v_0 &= (42.0\ut{m/s},60.0\degree)\\ t &= 5.50\ut{s}\\ \vec a &= -g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \vec v_0 &= (42.0\ut{m/s},60.0\degree)\\ &=21\i+21\sqrt{3}\j\ut{m/s}\\ \end{aligned}$$ (a) $h=?$ $$\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}at^2,\\ h &= \(21\sqrt{3}\)(5.50)+\frac{1}{2}(-g)(5.50)^2\\ &=\frac{231 }{2}\sqrt{3}-\frac{121 }{..